It's a matter of convention.
For many authors, "rings" are required to be unital rings (have a multiplicative unit); when rings are required to be unital, it makes sense to require the homomorphisms to be unital as well (viewing rings as general algebras with two binary, $+$ and $\times$, one unary, $-$, and two nullary operations, $0$ and $1$, homomorphisms are required to respect all the operations). This is the convention followed, for example, by Lam in his A First Course in Noncommutative Rings (to give a highly regarded, professional ring theorist example of someone who would agree with Wikipedia).
Another way to justify this is to recall that a monoid homomorphism is not merely a semigroup homomorphism between monoids: if $M$ and $N$ are monoids, a monoid homomorphism is a map $f\colon M\to N$ such that $f(ab)=f(a)f(b)$ and $f(e_M) = e_N$ holds. Thus, for instance, the map $(\mathbb{N},\times)\to (\mathbb{N}\times\mathbb{N},\cdot)$, where $\mathbb{N}$ are the nonnegative integers under multiplciation, and $\cdot$ is the coordinatewise multiplication, given by $a\mapsto (a,0)$, is not a monoid homomorphism, even though it is a semigroup homomorphism. (In a sense, it is a "happy accident" that any semigroup homomorphism between groups is also a group homomorphism; but it should really be defined as requiring that it map inverses to inverses and the identity to the identity).
If you view a ring as a set that has a structure of an abelian group under $+$ and a monoid under $\times$, with the two structures connected via the distributive laws, then it makes sense to require the homomorphisms between rings to simultaneously be group homomorphisms of the additive structure, and monoid homomorphisms of the multiplicative structure... and this requires the homomorphisms to map $1$ to $1$.
Under these requirements, the only time that the zero map can be a ring homomorphisms $\zeta\colon R\to S$ is when $S=\{0\}$ is the trivial ring.
For other authors, rings are not required to be unital; when the rings are not required to be unital, you certainly cannot expect homomorphisms to be unital. In that case, the zero map is always a homomorphism between two rings. This is the convention followed, for example, by ring theorists who do radical theory.
Best Answer
In general, given any sets $X$ and $Y$, a subset $S\subseteq X$, and a function $f:X\to Y$, the function $f|_S:S\to Y$ is the function $f$, where the domain is now considered to be $S$. The subset $S\subset R[x]$ consisting of constant polynomials can be identified with $R$ in the obvious manner, and in fact we usually write $R\subset R[x]$, leaving this identification implicit. So you are correct, $\phi|_R$ means exactly the restriction of $\phi:R[x]\to R$ to the subset of $R[x]$ consisting of constant polynomials.
Hint on how to proceed: show that, for any ring homomorphism $\psi:R[x]\to T$ where $T$ could be any ring whatsoever, the place where any $p\in R[x]$ is sent by $\psi$, namely $\psi(p)$, is determined completely by where elements of $R$ are sent, and where $x$ is sent. That is, if I tell you how $\psi$ acts on elements of $R$, and how $\psi$ acts on $x$, and tell you that $\psi$ is a ring homomorphism, you can figure out what $\psi$ does to any element of $R[x]$.
Note that we are assuming $\phi:R[x]\to R$ doesn't change elements of $R$, and then think about where $x$ is sent under the substitution homomorphism $\phi_a$...